a cart of weight 25N is released at the top of inclined plane of length 1m, which makes an angle of 30° with the ground .it rolls down the plane and hits another cart of weight 40N at the bottom of the inclined plane. calculate the speed of first cart at the bottom of the inclined and the speed at which both carts move together after the impact?

1 answer

h = 1*sin30 = = 0.5 m.

V1^2 = Vo^2 + 2g*h = 0 + 19.6*0.5 = 9.8
V1 = 3.13 m/s.

M1*g = 25N.
M1*9.8 = 25
M1 = 2.55 kg

M2*9.8 = 40N.
M2 = 4.08 kg

M1*V1 + M2*V2 = M1*V + M2*V
2.55*3.13 + 4.08*0 = 2.55*V + 4.08*V
7.982 = 6.63V
V = 1.20 m/s.